Решебник по алгебре 8 класс Мерзляк Задание 788

\[\boxed{\mathbf{788\ (788).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{60}{x} - \frac{60}{x + 10} = \frac{1}{5}\]

\[\frac{60}{x} - \frac{60}{x + 10} - \frac{1}{5} = 0;\ \ \ \ \ x \neq 0;\ \ \ \]

\[x \neq 5\]

\[- x^{2} - 10x + 300 = 0\]

\[x^{2} + 10x - 300 = 0\]

\[x_{1} + x_{2} = - 10,\ \ \]

\[x_{1}x_{2} = - 3000,\]

\[\text{\ \ }x_{1} = 50,\ \ x_{2} = - 60\]

\[Ответ:\ x = 50,\ x = - 60.\]

\[2)\ \frac{x}{x + 2} + \frac{x + 2}{x - 2} = \frac{16}{x^{2} - 4}\]

\[x^{2} - 2x + x^{2} + 4x + 4 - 16 = 0\]

\[x^{2} + x - 6 = 0\]

\[x_{1} + x_{2} = - 1,\ \ x_{1}x_{2} = - 6,\ \ \]

\[x_{1} = - 3,\ \ \]

\[x_{2} = 2\ (не\ подходит)\]

\[Ответ:\ x = - 3.\]

\[3)\ \frac{9}{x + 3} + \frac{14}{x - 3} = \frac{24}{x}\]

\[\ \frac{9}{x + 3} + \frac{14}{x - 3} - \frac{24}{x} = 0;\ \ \ \ x \neq 0;\]

\[x \neq 3;\ \ \ x \neq - 3\]

\[- x^{2} + 15x + 216 = 0\]

\[x^{2} - 15x - 216 = 0\]

\[x_{1} + x_{2} = 15,\ \ x_{1}x_{2} = - 216,\ \]

\[\ x_{1} = 24,\ \ x_{2} = - 9\]

\[Ответ:\ x = - 9;x = 24.\]

\[4)\ \frac{2y + 3}{2y + 2} - \frac{y + 1}{2y - 2} + \frac{1}{y^{2} - 1} = 0\]

\[y^{2} - y - 2 = 0\]

\[y_{1} + y_{2} = 1,\ \ y_{1}y_{2} = - 2,\]

\[\text{\ \ }y_{1} = 2,\ \ \]

\[y_{2} = - 1\ (не\ подходит)\]

\[Ответ:y = 2.\]

\[5)\ \frac{3x}{x^{2} - 10x + 25} - \frac{x - 3}{x^{2} - 5x} = \frac{1}{x}\]

\[\frac{3x}{(x - 5)^{2}} - \frac{x - 3}{x(x - 5)} - \frac{1}{x} = 0;\ \ \ \ \]

\[x \neq 0;\ \ x \neq 5\]

\[x^{2} + 18x - 40 = 0\]

\[x_{1} + x_{2} = - 18,\ \ x_{1}x_{2} = - 40,\ \]

\[\ x_{1} = 2,\ \ x_{2} = - 20\]

\[Ответ:\ x = - 20;x = 2.\]

\[x^{2} - 25x + 150 = 0\]

\[x_{1} + x_{2} = 25,\ \ x_{1}x_{2} = 150,\ \]

\[\ x_{1} = 10\ (не\ подходит),\ \ \]

\[x_{2} = 15\]

\[Ответ:x = 15.\]